Bernoulli\'s Equation is essentially a statement of conservation of energy for a

Question

Bernoulli's Equation is essentially a statement of conservation of energy for a fluid. The pressure is related to the work and other two terms are related to the kinetic energy per unit volume and the potential energy per unit volume.

Given two points (Point 1 and Point 2) in a fluid, Bernoulli's Equation states:

P1 + (1/2)v12 + g h1 = P2 + (1/2)v22 + g h2

where P is the pressure, is the density of the fluid, v is the velocity of the fluid, and h is the height of the fluid relative to a reference level.

Suppose we have a fluid with a density of 1.04×103 m/s2 flowing through a pipe. Let's try using Bernoulli's Equation in a couple of different situations.

Suppose the pipe is horizontal. We can place the reference line along the center of the pipe. That means that h1 and h2 are zero. Bernoulli's Equation simplifies to:

P1 + (1/2)v12 = P2 + (1/2)v22

If the pressure at Point 1 is 1.48×105 Pa, the velocity at Point 1 is 2.43 m/s, and the velocity at Point 2 is 4.57 m/s, what is the pressure at Point 2?

P1 + g h1 = P2 + g h2On the other hand, suppose the pipe moves from a height of 0.00 m at Point 1 to a height of 2.19 m at Point 2, but since the area of the pipe doesn't change, neither does the velocity. If we have the same velocity, we can subtract the kinetic energy from each side, leaving us with:

If the pressure at Point 1 is 1.48×105 Pa, what is the pressure at Point 2?

Explanation / Answer

when h1 = h2 = 0

we have the equation

P1 + 0.5 rho v1^2 = P2 + 0.5 rho v2^2

P2 = 1.48 *10^5 + 0.5* 1040 * (4.57^2 - 2.43^2)

P2 = 1.55 e 5 Pa
--------------------------------

use again P2 = P1 + rho g (h2-h1)

P2 = 1.48 e 5 + ( 1040 * 9.8 *(2.19-0)

P2 = 1.62 e 5 Pa

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